3 Statically Indeterminate Structures Reactions and internal forces cannot be found by statics alone (more unknown forces than independent equations of equilibrium). Results are dependent on the material from which the structure has been made. R C a unknown forces Only 1 useful equation of equilibrium b R + R 0 R Need to find another equation 3

4 R Flexibility or Force Method 1 a R static redundant C + b R 1 released structure Equation of compatibility expresses the fact that the change in length of the bar must be compatible with the conditions at the supports

5 Write the force-displacement relations. These take the mechanical properties of the material into account. b and E 1 R E Substituting into the equation of compatibility gives: R b R + 0 E E b *Note that flexibilities (b/e) and (/E) appear in this equation. Hence, this approach is called the flexibility method. Substituting into the equilibrium equation gives: R R a *Note that we have solved for forces. Hence, this approach is also called the force method. 5

7 Write the force-displacement relations and solve for the forces. R C1 Ra E C1E a and and C R Rb E CE b Substituting into the equilibrium equation gives: E a E b C 1 C + *Note that stiffnesses (E/a) and (E/b) appear in this equation. Hence, this approach is called the stiffness method. Using the compatibility condition (displacements equal) gives: ab E a C ab ( + b) E *Note that we have solved for displacement. Hence, this approach is also called the displacement method. 7

8 Finally, substituting into the expressions for forces gives: R R ab E ab E E a E b b a So, both the flexibility method and the stiffness method give the same result. The choice of approach will depend on the problem being solved. 8

12 Force-displacement relations TCDCD and E CD C TC E C Using the equilibrium moment equation, the compatibility equation, and the force-displacement relations, it is possible to solve for the forces in the wires. We find that T C 1.15 and T CD

17 olts and Turnbuckles The simplest way to produce a change in length. Nut and olt Distance travelled by the nut n p n number of turns (not necessarily an integer) p pitch of the screw (units mm / turn) Turnbuckle Right-hand screw eft-hand screw Distance travelled n p Often used to tension cables 17

18 ased on Gere, Example -9 Steel cable Turnbuckle Rigid plate Copper tube The slack is removed from the cables by rotating the turnbuckles until the assembly is snug but with no initial stresses (do not want to stretch the cables and compress the tube). Find the forces in the tube and cables when the turnbuckles are tightened by n turns, and determine the shortening of the tube. 18

19 If the turnbuckles are rotated through n turns, the cables will shorten by a distance 1 n p s c s The tensile forces in the cables s and the compressive force in the tube c must be such that the final lengths of the cables and tube is the same. 19

20 Equilibrium (forces must balance) s 0 c Compatibility (shortening of tube must equal shortening of cable) 3 1 Force-displacement 1 np s E 3 s c s c E c With these equations, we can solve for the forces in the tube and cables and for the shortening of the tube. 0

21 Temperature Effects Changes in temperature produce expansion or contraction of structural materials. When heated, the block expands in all three directions: x, y, z. For most structural materials, ε T α (ΔT) ε T thermal strain α coefficient of thermal expansion (HT, units 1/K or 1/ C) ΔT change in temperature The change in length of the block in NY direction can be found using T ε T α (ΔT), where is one of the block s dimensions. 1

23 Free expansion or contraction occurs when an object rests on a frictionless surface or hangs in open space. Then no stresses are produced by a uniform temperature change, but there are strains. In statically determinate structures, uniform temperature changes in the members produce thermal strains (and corresponding changes in length) without producing any corresponding stresses. statically indeterminate structure may or may not develop thermal stresses, depending on the character of the structure and the nature of the temperature changes. C D Since D can move horizontally, no stresses are developed when the entire structure is heated uniformly. If only some bars are heated, thermal stresses will develop. 3

26 Misfits and re-strains When a member is manufactured with a length slightly different from its prescribed length, the member will not fit into the structure as planned and the geometry of the structure will be different from what was planned. Such situations are misfits. Some misfits are created intentionally to introduce strains into the structure at the time it is built. ecause these strains exist before any loads are applied, they are called pre-strains. long with the pre-strains are usually pre-stresses. Examples: spokes in bicycle wheels, pre-tensioned faces of tennis racquets, shrink-fitted machine parts, pre-stressed concrete beams 6

27 If a structure is statically determinate, small misfits in one or more members will not produce strains or stresses, although the initial configuration will depart from the theoretical. Here, having CD longer than expected will not induce pre-strains or pre-stresses. can rotate to accommodate the length change. C D In a statically indeterminate structure, misfits cannot be accommodated without pre-stresses. 7

28 C E D F ssume that CD is slightly longer than prescribed. Then to assemble the structure, CD must be compressed by external forces (or EF must be stretched by external forces). The bars can then be fitted into place and the external loads released. s a result, the beam will deform and rotate. If CD is put into compression, EF will be in tension. So, pre-strains will exist and the structure will be pre-stressed, although no external loads are acting. When a load is added, additional stresses and strains will result. 8

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